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Quiz Chapter 8: Kinematics of Linear Motion

10 questions · Form 5 Additional Mathematics Bab 8: Kinematics of Linear Motion

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Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Given v = t² - 4t, find the displacement s in the first 3 seconds assuming s = 0 when t = 0.

  1. -9 m
  2. 9 m
  3. -3 m
  4. -18 m
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Answer: A

s = ∫ (t² - 4t) dt = [t³/3 - 2t²] from 0 to 3 = (273 - 2(9)) - 0 = 9 - 18 = -9 m.

2. Given v = 3t² - 12, find the displacement during the time interval 0 ≤ t ≤ 2 s.

  1. -16 m
  2. 16 m
  3. -24 m
  4. 8 m
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Answer: A

s = ∫₀² (3t² - 12) dt = [t³ - 12t]₀² = (8 - 24) - 0 = -16 m.

3. When an object moves with uniform velocity, what is its acceleration?

  1. a = 0
  2. a = v
  3. a > 0
  4. a = 9.8
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Answer: A

Uniform velocity means velocity is constant, so its derivative dv/dt = a = 0.

4. An object has a displacement function s = t³ - 6t² + 9t. Find its initial velocity.

  1. 9 ms⁻¹
  2. 0 ms⁻¹
  3. -6 ms⁻¹
  4. 3 ms⁻¹
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Answer: A

Velocity v = ds/dt = 3t² - 12t + 9. Initial velocity occurs at t = 0: v = 3(0)² - 12(0) + 9 = 9 ms⁻¹.

5. The acceleration of an object is given by a = 6t. If initial velocity is 0, find displacement s as a function of t (assuming s = 0 at t = 0).

  1. s = t³
  2. s = 3t²
  3. s = 6t³
  4. s = t³ / 3
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Answer: A

v = ∫ 6t dt = 3t² + c (since v(0)=0, c=0). s = ∫ 3t² dt = t³ + d (since s(0)=0, d=0). Thus, s = t³.

6. If velocity v = 6t - 18, at what time t does the particle reverse its direction of motion?

  1. t = 3 s
  2. t = 6 s
  3. t = 18 s
  4. t = 0 s
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Answer: A

Direction reverses when v = 0. Set 6t - 18 = 0 => 6t = 18 => t = 3 s.

7. Find the maximum velocity of a particle whose velocity function is v = 12t - 3t².

  1. 12 ms⁻¹
  2. 6 ms⁻¹
  3. 16 ms⁻¹
  4. 24 ms⁻¹
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Answer: A

Maximum velocity occurs when a = dv/dt = 0 => 12 - 6t = 0 => t = 2. Substitute t = 2 into v: v = 12(2) - 3(2)² = 24 - 12 = 12 ms⁻¹.

8. An object passes through fixed point O when t = 0. What is its displacement at t = 0?

  1. s = 0
  2. s = v
  3. s = 1
  4. s = a
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Answer: A

By definition, the fixed reference point O has displacement s = 0.

9. Which mathematical operation is used to find displacement s from velocity v?

  1. Integration of v with respect to t
  2. Differentiation of v with respect to t
  3. Dividing v by time t
  4. Squaring velocity v
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Answer: A

Since v = ds/dt, integrating v gives displacement s = ∫ v dt.

10. Find the acceleration of a particle at t = 2 s if its velocity is v = 2t³ - 5t.

  1. 19 ms⁻²
  2. 24 ms⁻²
  3. 11 ms⁻²
  4. 16 ms⁻²
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Answer: A

a = dv/dt = 6t² - 5. At t = 2: a = 6(2)² - 5 = 24 - 5 = 19 ms⁻².

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